Let E be a point with triangles DBA and EDC are congruence(AB=CD,AD=CE,ang(BAD)=x=ang(DCE), ang(ACE)=4x). Then DB=DE and ang(EDC)=2x and henceang(EBD)=x=ang(BED).Furthermore, since BA=CD, BE=EC=AD=CA and ang(EBA)=3x=ang(ACD), triangles BAE and CDA are congruence. we see that AE=AD and so triangle AEC is an equilateral. we get 4x=60 degree, that is, x=15 degree.

To Rengaraj PradheepKannahGreetings.In your solution of problem 6, there’s something I didn’t understand.The first circle is defined as having center at the midpoint G of AB (with diameter AB) and the second circle has center at the midpoint E of CD (with diameter CD).I can’t see why the first circle meets CD in its midpoint E. If we don’t assure that E lies on the first circle, it’s not clear that AG = GE.

To Nilton LapaGreetings to you too.Did you notice DBC is an Isosceles triangle.Since E is the Mid point of DC BEC is 90 Deg.As AB is the diameter of the first circle according to the angle on semi circle is 90 DegThe first circle should go thriugh E.

Let E be a point on the extension AB such that AE=AC. Bisect ∠A and let a point F on the angle bisector of ∠A such that m∠AEF=m∠ACF=x. Then ∆AFE≅∆AFC by ASA and at the same time they are both isosceles. Let a point H on BC such that the angle bisector of ∠A intersect BC at H. Then m∠CFH=m∠EFH=m∠FEH=m∠FCG=2x, also m∠EBC=5x and m∠EGC=6x. An exterior angle is equal to the sum of its remote interior angles. Again by ASA, ∆EHF≅∆CHF and they are also isosceles triangles, then EH=FH=CH. By perpendicular bisector concurrence theorem, CG⊥EF that makes ∠EGC a right angle. Therefore: 6x=90°,x=15°